Question

Calculate the standard enthalpy of formation of CH₃OH_(I) from the following data:

(i) CH₃OH_(I) + 3/2 O_2(g) → CO_2(g) + 2H₃O_(I) ∆_r H^⊖= -726 kJ mol^-1

(ii) C_(s) + O_2(g) → CO_2(g) ; ∆_CH^⊖ = – 393 kJ mol^-1 

(iii) H_2(g) + 1/2 O_2(g) → H₂O_(I) ∆_fH^⊖ = – 286 kJ mol^-1

Answer 1

CH₃OH_(I) + 3/2 O_2(g) → CO_2(g) + 2H₃O_(I) ∆_r H^⊖= -726 kJ mol^-1.....(i)

C_(s) + O_2(g) → CO_2(g) ; ∆_CH^⊖ = – 393 kJ mol^-1 .....(ii)

H_2(g) + 1/2 O_2(g) → H₂O_(I) ∆_fH^⊖ = – 286 kJ mol^-1 ...(iii)

The equation we aim at;

C_(s) + 2H_2(g) + O_2(g) → CH₃OH_(I) ∆f H^⊖ = \(\pm\) ?...(iv)